我有这个:
res.list<-list(
list(
structure(list(A = 1L, B = "X2", D = 1L), class = "data.frame", row.names = c(NA, -1L)),
structure(list(A = 1L, B = "X3, X14, X17", D = 3L), class = "data.frame", row.names = c(NA, -1L)),
structure(list(A = 1L, B = "X4, X14, X17", D = 3L), class = "data.frame", row.names = c(NA, -1L)))
,
list(
structure(list(A = 2L, B = "X17, X19", D = 2L), class = "data.frame", row.names = c(NA, -1L)),
structure(list(A = 2L, B = "X2, X17, X19", D = 3L), class = "data.frame", row.names = c(NA, -1L)),
structure(list(A = 2L, B = "X3, X17, X19", D = 3L), class = "data.frame", row.names = c(NA, -1L))))
做下面的程序最有效和通用的方法是什么?
rbind(do.call(rbind.data.frame,res.list[[1]]),
do.call(rbind.data.frame,res.list[[2]]))
问好;
使用bind_rows
可能更容易
library(dplyr)
bind_rows(res.list)
A B D
1 1 X2 1
2 1 X3, X14, X17 3
3 1 X4, X14, X17 3
4 2 X17, X19 2
5 2 X2, X17, X19 3
6 2 X3, X17, X19 3